A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

if Given E=mc^2 C representing the constant Distance/Time. What if Time is then considered to actually be a single local computation cycle. But the computational cycle is not constant throughout the universe, and is made longer by computational complexity. So modified by the variable x say, which represents an extension to the clock cycle, caused by compleXity. Local space then is something like a fixed clock computer, that gets slowed down by extra jobs added to it, thus slowing all its jobs. And we get E=m(d/(t+x))^2. Now if x increases, the other variables must change. Any thoughts?

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Could this equation describe what we observe?

  2. pooja195
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @perl @jigglypuff314

  3. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    C represents the speed of light, its a constant

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    true, but Im saying, what if we changed that. and said.. it isn't constant.. and we modified it.. to include an interaction between itself and everything that is going on around it.. Took away some computational energy to deal with an increase in information exchange and complexity. And then compare that to what we observe when it comes to black holes, and universal expansion. I was just curious to see what people a lot smarter than I am, would think of that idea.

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah well.. I guess back to the business of learning math...

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hence the proposed. E=m(d/(t+x))^2

  7. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Interesting idea, unfortunately this is outside my scope of knowledge. I will tag others :) @Michele_Laino

  8. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Thinking outside the box is a good skill.

  9. nincompoop
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes. It describes what we observe.

  10. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    your formula: E=m(d/(t+x))^2 was correct if we deal with Classical Mechanics. When we have to deal with Relativistic Mechanics, we have to do another reasoning. The starting point is that the speed of a particle, with respect to a inertial system, can not be greater or equal to speed light c When a constant force is acting on a particle, its speed can not reach the constant C (speed light), so where it ends the extra energy that our particle receives from the work done by that force? The answer is: that extra energy which our particle receives from that external force, will transform itself in extra mass of our particle. In other words as the speed of our particle approaches to C, its inertia increases, according to this relation ship: \[\Large m = \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \gamma {m_0}\] where m_0 is the mass of our particle, measured in a inertial system in which it is at rest.

  11. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    from such reasoning we are led to the correct meaning of the famous Einstein's relationship: \[\Large E = \Delta m \times {c^2}\]

  12. nincompoop
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    relativity intro

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome .. thanks for sharing.

  14. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks! :)

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.